IMPULSE AND MOMENTUM 243
target and the bullet a common velocity v'. Therefore if d! be the new depth of penetration we have
Fd' = i mv* - J (m + M) v'2, (2)
where M is the mass of the target. Eliminating F between equations (1) and (2) we get
But by the conservation of momentum we have
mv — (m 4- M) vf. . (4)
Therefore eliminating the velocities between equations (3) and (4) we get
M + m
It is evident from equation (5) that when the target is free, but very large compared with the bullet, the depth penetrated is about the same as when it is fixed.
1. A particle which weighs 2 ounces describes a circle of 1.5 feet radius on a smooth horizontal table. If it makes one complete revolution in every 3 seconds find the magnitude and direction of the impulse imparted by the force, which keeps the particle in the circle,
(a) in one-quarter of a revolution;
(b) in one-half of a revolution;
(c) in three-quarters of a revolution;
(d) in one complete revolution.
2. Find the expression for the impulse imparted to a particle in describing an arc of a circle with uniform speed.
3. Considering the rate of change of the momentum of a particle which describes a uniform circular motion derive the expression for the central force.
4. If we neglect the resistance of the air to the motion of a projectile what can we state with regard to the components of the momentum in the horizontal and vertical directions?
5. A train which weighs 100 tons runs due south at the rate of one mile a minute. Find the lateral force on the western rails due to the rotation of the earth, while the train passes the line of 30° latitude.
6. At what latitude will the force of the preceding problem be a maximum? Determine its amount.